Steiner Roman surface
The Roman surface is a projective plane. To see this we first translate the surface to a sphere (see for instance http://mathworld.wolfram.com/RomanSurface.html):
On the sphere two elliptic bundles are drawn and the same bundles on the roman surface. The 3 violet great circles on the sphere correspond to the 3 axes of the roman surface.
A complete quadrilateral on the projective plane corresponds to the 4 circles below and the 3 diagonals of this quadrilateral are the 3 straight lines of self intersection.
If the projective plane is translated to a sphere whose antipodal points are identified (elliptic model) we get a (spherical) cuboctahedron for the quadrilateral and a (spherical) octahedron for its diagonals:
The 6 double points i, j, k, I, J and K become 6 antipodal pairs of vertices of the cuboctahedron and the central triple point E becomes the 3 pairs of antipodal vertices of the octahedron. After a transformation of the sphere S3 that has this sphere S2 as 3-dim section, we make a stereographic projection from the 'south pole' with (t,x,y,z)=(-1,0,0,0):
Making a projective plane:
Getting back the third violet diagonal of the complete quadrilateral by changing the affine choice of the line at infinity:
The complete quadrilateral jKJk is a coassociative square and iEI the complementary associative triple. Likewise jEJ and iKJk are complementary and kEK and iJIj. Then there are 4 triangles in one color that make a coassociative square together with E: EIJK (black) with ijk as complementary associative triple, EiJk (dark blue) and IKj, EijK brown) and IJk, EIjk (green) and iJK.